Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.
Note: Do not modify the linked list.
Example 1:
Input: head = [3,2,0,-4], pos = 1
Output: tail connects to node index 1
Explanation: There is a cycle in the linked list, where tail connects to the second node.
Example 2:
Input: head = [1,2], pos = 0
Output: tail connects to node index 0
Explanation: There is a cycle in the linked list, where tail connects to the first node.
Example 3:
Input: head = [1], pos = -1
Output: no cycle
Explanation: There is no cycle in the linked list.
Follow up:
Can you solve it without using extra space?
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这个题是不仅是判断链表中是否存在环，还要返回环的开始的位置。
用双指针，与141. Linked List Cycle 的类似，关于怎么返回这个位置，可以参考这个大佬的博客：http://www.cnblogs.com/hiddenfox/p/3408931.html
C++代码：
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *detectCycle(ListNode *head) {
ListNode *slow,*fast;
slow = head;
fast = head;
while(true){
if(fast == NULL || fast->next == NULL)
return NULL;
slow = slow->next;
fast = fast->next->next;
if(slow == fast)
break;
}
slow = head;
while(slow != fast){
slow = slow->next;
fast = fast->next;
}
return slow;
}
};